Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Convergence of series, sequences and functions and different modes of convergence.
1
vote
Accepted
How to prove this iterative convergence of trigonometric functions
Let's use induction on all sequences of length $n$.
For length $1$ we have $2\sin(\frac{\pi}4 s_0) = s_0$, but actually $2\sin(\frac{\pi}4 s_0) = s_0 \sqrt 2$, so there's a slight error in the questio …
7
votes
$\lim_{n \to \infty} \frac{2^n}{n} \left[ 1 - \sum_{k = 1}^{n-1} \frac{(1 - \lambda 2^{-n})^...
We have $$(1 - \lambda 2^{-n})^{2^k} = e ^ {2^k \log(1 - \lambda 2^{-n})} = e ^ {2^k ( - \lambda 2^{-n} + O_\lambda(4^{-n}))} = e^{-\lambda 2^{k-n} + O_\lambda(2^{-n})} = (1 + O_\lambda(2^{-n})) e^{-\ …
46
votes
Accepted
A challenging (for me) limit calculation
This limit converges to $\frac{\sqrt3}2$. The idea is that $\sin(x) = x - \frac{x^3}6 + O(x^5)$, so we start with $\frac1{\sqrt n}$ and repeatedly subtract $\frac{x^3}6$. We can approximate this discr …